I've done a simple-minded transliteration of your code into code using nested
lapply's. I doubt that it buys you much in terms of performance (or even
clarity, which is really one of the main advantages of the `apply` family).
> A
[,1] [,2] [,3] [,4] [,5]
[1,] 3.06097 6.507521 10.99610 12.05556 15.10388
[2,] Inf 11.818495 15.85044 16.69465 19.70425
[3,] Inf Inf Inf 19.14779 22.30343
[4,] Inf Inf Inf Inf 26.11170
[5,] Inf Inf Inf Inf 28.29882
> B
[,1] [,2] [,3] [,4] [,5]
[1,] 3.06097 6.507521 10.99610 12.05556 15.10388
[2,] Inf 11.818495 15.85044 16.69465 19.70425
[3,] Inf Inf Inf 19.14779 22.30343
[4,] Inf Inf Inf Inf 26.11170
[5,] Inf Inf Inf Inf 28.29882
> all.equal(A, B)
[1] TRUE
If I happen to think of a more-elegant approach, I'll let you know.
-- Mike
Appendix: code
==============
###### Anne's code
getResult <- function(d) {
#examplefunction
weighted.mean(x=d[,1], w=d[,2])
}
#example data setup
n=20;
set.seed(1)
g=rep(1:5,each=4)
df=as.data.frame(cbind( sort(rnorm(mean=15,sd=10, n)),runif(n), rbinom(n, 1,
0.4) , g )); df
getResult(df)
i0=c(1,2,4,5,5)
ng= length(unique(g))
#initiation of result matrix
A=matrix(Inf, ng, ng); A
for(i in 1:ng)
{ cat("i:",i,"")
for(j in i0[i]:ng) {
ok= !is.na(match(g,i:j)); cat("j:",j,"\n");
A[i,j]=getResult(d=df[ok,])
} #endfor (j)
} #end for (i)
A
###### Mike's code
n <- 20;
set.seed(1)
g <- rep(1:5,each=4)
df <- as.data.frame(cbind(sort(rnorm(mean=15,sd=10, n)),
runif(n),
rbinom(n, 1, 0.4),
g )); df
getResult(df)
i0 <- c(1,2,4,5,5)
ng <- length(unique(g))
B <- matrix(Inf, ng, ng);
invisible(lapply(1:ng, function(i) {
lapply(i0[i]:ng, function(j) {
ok <- !is.na(match(g, i:j))
B[i, j] <<- getResult(df[ok, ])
})
}))
B
all.equal(A, B)
On Mon, Oct 12, 2015 at 5:55 PM, Annie Hawk via R-help
<r-help@r-project.org> wrote:
> HI R-experts,
>
>
> I am trying to speed up my calculation of the A results below and replace the
> for loop withsome functionals like lapply. After manyreadings, trial and
> error, I still have no success. Would anyone please give me some hints
> onthat?
>
> Thank you in advance.
>
> Anne
>
>
> The program is this, I have a complicated function and itneeds to operate on
> some subsets of a dataset many times, depending on thevalues of group. I
> simplify the functionand dataset for this example run.
>
> getResult <- function(d) {
>
> #examplefunction
>
> weighted.mean(x=d[,1], w=d[,2])
>
> }
>
>
>
> #example data setup
>
> n=20;
>
> set.seed(1)
>
> g=rep(1:5,each=4)
>
> df=as.data.frame(cbind( sort(rnorm(mean=15,sd=10, n)),runif(n), rbinom(n, 1,
> 0.4) , g )); df
>
> getResult(df)
>
> i0=c(1,2,4,5,5)
>
> ng= length(unique(g))
>
>
>
> #initiation of result matrix
>
> A=matrix(Inf, ng, ng); A
>
> for(i in 1:ng)
>
> { cat("i:",i,"")
>
> for(jin i0[i]:ng) {
>
> ok= !is.na(match(g,i:j)); cat("j:",j,"\n");
>
> A[i,j]=getResult(d=df[ok,])
>
> } #endfor (j)
>
> } #end for (i)
>
> Is there an elegant way to remove the for loop here? I try to make it flat
> for faster run but Icannot figure out how to subset the observations faster
> without error to apply the functiongetResult. Any hint is appreciated.
>
>
>
>
>
> on another note, is there a more elegant way to initiate the list as follows?
>
> mylist=list(); w=rep(4,5)
>
> for (i in 1:5) mylist[[i]]=w[i:5]
>
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
